The following equation is true for all real values of $z$ for which the expression on the left is defined, and $C$ is a polynomial expression. $\dfrac{z^2-16z+64}{40-5z} \cdot \dfrac{C}{z^2-10z+16}=1$ What is $C$ ? $C=$
Answer: The left side of the equation is a product of two rational expressions, while the right side is simply $1$. This means that the numerator and the denominator of the resulting product on the left side should cancel out completely. In order to solve for $C$, let's multiply the expressions and simplify as much as we can. We start with factoring the numerators and the denominators of the two expressions. [Why are we doing this?] The numerator, $z^2-16z+64$, of the first expression can be factored as $(z-8)(z-8)$ using the perfect square pattern. The denominator, $40-5z$, of the first expression can be factored as $5(8-z)$ by factoring out $5$. The denominator, $z^2-10z+16$, of the second expression can be factored as $(z-8)(z-2)$ by using the sum-product pattern. Now the product looks as follows: $\dfrac{(z-8)(z-8)}{5(8-z)} \cdot \dfrac{C}{(z-8)(z-2)}$ To find the product of two rational expressions, we multiply across, then simplify: [What's that?] $\begin{aligned} &\phantom{=} \dfrac{(z-8)(z-8)}{5(8-z)} \cdot \dfrac{C}{(z-8)(z-2)} \\\\\\ &= \dfrac{(z-8)(z-8) \cdot C}{5(8-z) \cdot (z-8)(z-2)} &\text{Multiply across.}\\\\\\ &= \dfrac{{\cancel{(z-8)}}{\cancel{(z-8)}} C}{5(-1)\!\!{\cancel{(z-8)}}{\cancel{ (z-8)}}(z-2)} &\text{Cancel out common factors.}\\\\\\\\ &=\dfrac{C}{-5(z-2)} \end{aligned}$ After this simplification, our equation now looks like: $\dfrac{C}{-5(z-2)}=1$ We can conclude that the numerator and the denominator of the expression on the left must be equal. [Why?] In other words, $C=-5(z-2)$, which is equivalent to $10-5z$.